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5.8 Some Applications of Algebraic Computing  229


                          H 2  H 4        h 2  h 4
                                   H 5            h 5
                          H 1  H 3     = h 1  h 3
                                                        ,
                                   H 4            h 4
                            1  H 2  H 3        1  h 2  h 3

                                           h 1  h 3  h 4  h 5

                                           1

                               H 3            h 2  h 3  h 4

                                                           .         (5.8.2)
                                   H 5
                                       =
                               H 1  H 3       h 1  h 2

                                                      h 3
                                                   1  h 1

          5.8.3  Hankel Determinants with Hankel Elements
          Let
                           A n = |φ r+m | n ,  0 ≤ m ≤ 2n − 2,       (5.8.3)
          which is an Hankelian (or a Turanian).
            Let
                                 B r = A 2


                                    =     φ r  φ r+1    .            (5.8.4)
                                       φ r+1  φ r+2

          Then B r , B r+1 , and B r+2 are each Hankelians of order 2 and are each
          minors of A 3 :
                                          (3)
                                   B r = A  ,
                                          33
                                          (3)   (3)
                                 B r+1 = A  = A   ,
                                          31    13
                                          (3)
                                 B r+2 = A 11  .                     (5.8.5)
          Hence, applying the Jacobi identity (Section 3.6),
                                               (3)  (3)
                              B r+2            11
                                               A   A
                                    B r+1    =     (3)  13
                              B r+1          A     A
                                                    (3)
                                               31   33
                                     B r
                                                (3)
                                          = A 3 A
                                                13,13
                                          = φ 2 A 3 .                (5.8.6)
            Now redefine B r . Let B r = A 3 . Then, B r , B r+1 ,...,B r+4 are each second
          minors of A 5 :
                                    (5)
                             B r = A    ,
                                    45,45
                                      (5)      (5)
                           B r+1 = −A     = −A     ,
                                      15,45    45,15
                                    (5)     (5)     (5)
                           B r+2 = A    = A     = A     ,
                                    12,45   15,15   45,12
                                      (5)      (5)
                           B r+3 = −A 12,15  = −A 15,12 ,
                                    (5)
                           B r+4 = A    .                            (5.8.7)
                                    12,12
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